PAT A1140

最新推荐文章于 2021-04-17 10:35:05 发布

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本文详细解析了一种特殊的整数序列——看说序列（Look-and-say Sequence）的生成算法。通过统计元素连续出现的个数，并以（元素：个数）的形式输出，实现了对任意给定数字D的N次看说序列的计算。

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1140 Look-and-say Sequence （20 分）
Look-and-say sequence is a sequence of integers as the following:

D, D1, D111, D113, D11231, D112213111, …
where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is one D in the 1st number, and hence it is D1; the 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to 11), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, two 1’s, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digit D.

Input Specification:
Each input file contains one test case, which gives D (in [0, 9]) and a positive integer N (≤ 40), separated by a space.

Output Specification:
Print in a line the Nth number in a look-and-say sequence of D.

Sample Input:
1 8
Sample Output:
1123123111
开始其实不是很懂什么意思，题意其实就是统计元素连续个数并以（元素：个数）的形式输出
如
1
11
12
1121
122111
112213
12221131
1123123111

#include<cstdio>
#include<vector>
#include<string>
#include<iostream>
using namespace std;

int main() {
	int D, n;
	scanf("%d%d", &D, &n);
	int temp;
	int count;
	string ans = to_string(D);
	for (int i = 1; i < n; i++) {
		string t;
		int j = 0;
		while(j < ans.size()){
			count = 0;
			temp = -1;
			for (int k = j; k < ans.size(); k++) {
				if (ans[k] == ans[j]) {
					count++;
				}
				else {
					temp = k;
					break;
				}
			}
			t += ans[j] + to_string(count);
			j = temp;
		}
		ans = t;
	}
	cout << ans;
	return 0;
}